After a typical ultrahard Sysudoku Basic,

Sudokuwiki starts with an APE so unusual as to change my working definition of the APE. APE stands for Aligned Pair Elimination. From every example so far, I thought it was the two elimination cells that were aligned in a row or column. No, it’s that two cells are simultaneously aligned with ALS that can prohibit any combination of two candidates, one from each cell, that see two of the ALS’ value groups. If every combination of a candidate in one of the cells is so prohibited, it is false.

Here, every combination of 7r9c7 with 2, 4 and 9 in r7c1 is prohibited. The blue ALS stops 27 and 47. The red ALS stops 27 and 29. One way to avoid repeating work is to list every combination of the two cells, then cross out combinations stopped by the ALS seeing both cells. There are 4 combinations of 2 with other values, and 10 combinations of other values with other values. A policy for a reasonable human search for APE is complicated.

Sudokuwiki, the computer code, has no such problem with APE. Four combinations to filter through the ALS this time.

It may be the number of ALS and cell pairs that’s more of a problem. How many combinations this time? Only 5.

Next, Sudokuwiki adds this AIC ANL with an ALS node.

Then it identifies 3r8c6 as a pattern orphan, without explaining why. We analyze the 3-panel, looking for a humanly possible way to do this.

The 3-panel reveals a limited number of West to East freeforms crossing r8c6: 3 from r2c1 and one from r3c1. Looking for a toxic orphan. Note that those visiting r8c6 must also visit r9c7. Among the X-panels, we don’t see another whose every pattern must visit these two cells.

But Beeby’s note identifies 9 as the restraining value. Of the five East to West freeform enumerated 9 patterns, three are stopped by 3r6c5 and the other two by 3r5c9. So 3r8c6 is a toxic orphan. After announcing the orphan, Sudokuwiki gives up. Switching to Beeby, we get the same result, but with the APE’s handled differently. You can’t prescribe candidates to either solver, only given clues, so we get to see Beeby redo the APE eliminations.

First, the 7r7c4 is an ALS_37.

Then 2r7c7 disappears via a grouped ALS_42.

The 7r2c4 removal takes an ALS wing, which can also be interpreted as an ANL with the slink chain with two or three ALS nodes.

Then after identifying the toxic orphan 3r8c6, all the Beeby methods fit for human computers are exhausted, and its trial time.

The Sysudoku bv scan reveals Single Alternate Sue de Coq NW3 = 9(6+7)(2+5) + 952. If the trial fails, the remaining SdC NWc3 = 9(6+7)(2+5) eliminates 6r2c1 and 6r3c1 in the NW box. We add a small cluster.

After the brief follow up

and a reload, Beeby continues the trial next week. The reload misses APE elimination 2r7c8, but that has no effect on the trial.

Here is the trial grid, after the removal of the toxic orphan last week.

With orange on trial, blue is confirmed quickly,

the hidden UR removes 2r9c2, and W6 removes the last 5 candidate in the West box, placing the red army,

, but leaving the blue green cluster unresolved,

The naked quad in r8 leaves SE1 and an expanded cluster trapping 4r6c7, and bringing

a naked triple in c6, and

another hidden unique rectangle, followed by

an ALS_85, with a SWc1 boxline removal.

Beeby now recycles the ALS to use its internal slink between two single value groups as the starting slink in a 1-way. 1r9c6 is made false directly if 1r7c6 is true, and by the AIC if it is false.

After (S1, S5) and the 9r7c1 expansion trap, Beeby launches another 1-way from a line slink, this time, 2r4c6. If the starter is false the chain removes the victim straight up.

Beeby lines up two ALS against the combinations of two cells in this APE. There’s no APE button but this is a Beeby ALS “other”.

After the follow up

traps 6r4c3,

blue is wrapped as 8r6c4 turns blue, and the grid turns green..

Next week, its Stefan Heine’s ultrahardcore 355.

Picking up from last post, Beeby now comes in with a complex 1-way. That’s how it starts, with a bv slink partner seeing other 3’s. It works out as a 1-way ANL with a branch coming off to whack 4r4c9 and create the necessary r4 slink.

Of course, human solvers don’t “spot” these. They spot possible beginnings and keep chains going, looking back for branches to create slinks.

The removal erases the r1 ALS 2356 on the ALS-map, but it leaves all the AS required for the ALS- wing that Beeby finds next. This gives you an opportunity to search for that ALS-wing on that map before looking. Go back and do that while I create a little diversion by discussing the trade-off between marking and not marking the multi-candidate value groups on the map. The markings would definitely aid you in building the inference chain, but would would add considerable unproductive clutter to the map.

Next is a Beeby ALS_wing, a 3 node ALS chain ANL. You would probably start this chain at the bv r5c5, noticing the winks to single value groups in two ALS on the map. That would lead you to discover that the shared value 8 and the slinks with the 8-groups and the entry candidates. One group is a box group, and the other is a line group.

Were there any other ALS wings?

Then Beeby uses a branch enhanced AIC for three eliminations: a complex 1-way from 8r8c1 to confirm 4r4c1, and extending it to confirm 3r2c1, and having removed two c1 8’s by 1-way logic, invoking the boxline SWc1 to remove 8r7c2.

Parts of that structure show up again in this complex 1-way from 2r2c6 which, true or false, extinguishes 2r2c1.

We hopefully start a second cluster as a simple 1-way removes 2r4c2 whether 2r1c2 is true or false.

Then comes the very complex 1-way, the AIC starting from 3r2c1 and sprouting four slink making branches.

Its removal of 3r2c9 allows this overlapped ALS_62.

Beeby now continues with a puzzling “complex” 1-way with an unnecessary forward branch. The Beeby annotation shows an ending slink to 7r3c7, which is valid, but unnecessary. The forward branch from 7r7c4 could be used to reinforce that the chain assumes 7r3c4 is false. It might be what Beeby does when the starting slink is a line slink, rather than a bv slink.

Sudokuwiki records this one as a digit forcing chain.

Now consider the labeling over abundance. This ALS_36 is also an ALS boomer from 3r7c7, and a unbranched ALS 1-way from 6r1c7. It depends on what you were looking for when you found it.

Wouldn’t you know it? Both ALS were added to the map.

If you can wade through one more ultra complex 1-way, we can pick off another 3 candidate. It starts at 3r2c1 again, and uses much or the same structure as the previous one.

Now the pattern analysis twist. The red/orange cluster expands with the 3 clue, and Sudokuwiki claims that 2r2c3 is an orphan.

Checking it out on the 2-panel, the North to South freeforms are restricted, but there are two patterns including 2r2c3. Note that an attempted freeform through r3c9 is stymied, because it leaves no entry into the East box. Now we know that patterns including 2r2c3 must also visit r3c4, r4c6 and r6c9.

If any combination of these three cells exclude all patterns of any value other than 2, 2r2c3 is an orphan!

Sudokuwiki notes don’t explain that reasoning, and enumerating all patterns of all other values is not normally feasible, but here we notice that 7-patterns are quite restricted, and that delivers the goods.

Going North, all 7 patterns visit r3c4, with one exception, the short-dashed one in the right panel, from r8c2. That pattern visits r6c9. So 2r2c3 excludes all 7 patterns, an unforgivable offense.

Beeby is silent here, but in a coming ultrahardcore in the review, it does acknowledge this kind of rationale for removing an orphan candidate. For future reference, we’ll call this removal method the *toxic orphan*.

Looking at the diagram above, there is a much easier way to eliminate the “orphan”, by a combination of coloring and the Sue de Coq. NEr2 contains 89(2+7) if red is true or 69(2+7) if it is orange. In either case, the matching 2 or 7 in the r2 remainder must come from r2c6.

We continue from here next week to a solution. That means you may need a preview of 355, the next review ultrahardcore.

Ultrahardcore 311 yields no clue in to the bypass, and doubles down in box marking, leaving basic with two naked singles, a naked pair, and a more typical number of slinks and bv.

The line marked grid is remarkably playable, considering the very long fill strings.

Sudokuwiki starts advanced with a grouped “digit forcing chain”, more accurately classified by Beeby as a 1-way from 4r3c6. . The starter sees the victim, and if true, it removes 4r4c6 directly. If 4r3c6 is false, the inference chain confirms 6r4c6, with the same effect.

Another dfc by Sudokuwiki notes. It is an AIC ANL, in Sysudoku Advanced sequence, and would come in AIC building. It could start at either terminal. For simpler graphics we omit the closing winks on the almost nice loops.

This grouped boomer from 4r3c6 implies N4. Beeby has already found N4 in basic, where explanations are not given. Want to go back to the line marked grid and explain how Beeby did that with basic operations? I missed it.

Another dfc AIC ANL, or what Beeby calls a “simple chain”.

Sudokuwiki finds this ALS_65! And then gives up.

The ALS spotting problem for the human solver is the large number of ALS, and the difficulty of seeing ALS_XZ pairs at once

A systematic approach is to maintain an ALS_XZ map showing only the ALS having two value groups that are single or aligned for possible Z and X with other ALS.

Here’s an ALS_XZ map at this point. Every ALS is contained in a line or box so three or four colors are enough to distinguish them within their unit.

The Z value groups in each ALS must see each other completely. The X value groups must be seen completely by some outside candidate.

In r1, no curves are drawn for r1c47 ALS 236 or r1c457 ALS 2356. Each as a Z=6, but no X restricted common with r8c1579 ALS14568.

Instead of redrawing the ALS_XZ map on each move, a more practical approach is to bring the updated ALS_XZ slide along as you solve, updating it on each move, and keeping an ALS on the map as long as a potential Z and X value groups are available.

Sudokuwiki found single 5 of r1c247 ALS 2356. Its ALS XZ algorithm must do some kind of limited search. Beeby’s ALS search is more exhaustive.

Next week, we continue from here with Beeby on this long November march.

]]>The later right pages of ultrahardcore review have become tougher on the bypass. And line marking doesn’t usually start with 5f: lines. Fortunately, a hidden pair clears a little of the congestion

Advanced methods start with a 5-chain ANL, by both solvers, and a Sudokuwiki 597-wing.

Beeby does XYZ-wings as ALS_XZ. Here’s the ALS_57.

Beeby’s next step is an ALS full Nelson, its ALS Wing It’s not the imbedded AIC that’s hard to spot, its the 3 or ore ALS with single or aligned value groups.

Sudokuwiki does it as a Death Blossom. The victims grab 9’s from both ALS which in turn grab both values from the stem r2c6.

Beeby flashes its ALS advantage by going one ALS value group better in the same ALS for clue C3

The assault on the 9’s continues with an AIC boomer from 2r7c5 and a unque rectangle, type 4. In Beeby notes the boomer is a simple discontinuous loop, implying a unforced 1-way from 9r6c5

The next Beeby ALS wing is a ANL with ALS value group of four candidates. How do you define the problem to have the code look for such a thing?

The two solvers push their bailout buttons after one more jab at ultrahardcore 267. Beeby ends with an ordinary 1-way in which finds 4r6 is false when it sees a true 4r6c4, when a false 4r6c4 starts an AIC making it false.

The Sudokuwiki contribution is the 5-pattern orphan 5r1c1.

Here is the 5 panel showing many West to East freeforms from 5r3c1 and the failure of any from 5r1c1. Reason through the why not and you’re onto patterns by freeform.

The solvers having had their say and not much to color, I’m on the hunt for a trial. And theres a possible Single Alternate Sue de Coq at Wr5 as marked above. The contents formula is

Wr5 = 8(2+4)(7+9) + [7(89+98) +897 ]

That second term is a bit awkward to test. The progress already made on r5, and tracing technology, gives us a way out. Either Wr5 = 8(2+4)(7+9) with 2 or 4 present, or if 2 and 4 are both absent, then in r5, Wr5 is a naked triple, r5c6 is 5, r5c9 is 4 and r5c4 is 2. A trial failure doesn’t gain much, only a 4r5c9 removal, but given the investment so far, we give it a shot. And wouldn’t you know it, this time, we reach no contradiction, only the solution.

Next week, we start ultrahardcore 311.

]]>Beeby does an ALS_12 with an aligned Z value group in r1 and a single Z value in the green ALS. In a DIY hidden unique rectangle, were 3r3c5 to be true, so would the adjacent corners be 8. Then 8r3c6 forces 7r8c6, a value interchangeable rectangle, and an obvious double solution.

Continuing on a second 2-chain ANL we believe this may end, after all.

Beeby finds two ALS-wings, black and red, packaged in three ALS. This remarkable structure raises the prospect that other ALS aided ANL are invoked in the Beeby solver, beyond ALS wings.

Beeby follows this with two more ALS-wings. Here, one ANL terminal ALS value group is a single candidate, and the other terminal ALS value group is a box group. In the previous one, all terminal value groups were line groups.

In a surprising remap of the grid, a four member ALS-chain removes 3r3c3, with ANL terminals being a bv candidate and a box group. The removal triggers a Wc3 boxline.

Cluster expansions now show a merge:

red =>blue and orange => green.

The two clusters merge

Beeby doesn’t do color logic. On the merged grid Beeby finds this removal as a discontinuous loop, from 5r5c5. In coloring it’s a regular trap. 5r5c7 is false regardless of which color is true. If blue, it sees a blue 5. If green it loses its cell to green 8.

The spotting rule is seeing one color and living with the other, out you go. A bit like marriage. As 6r5c7 turns blue, 6r5c9 is trapped.

With this extensive cluster, a color trial would resolve 223 immediately. Beeby brings back a variation of that four node ALS-wing, and a NWr1 boxline. Actually, you can leave out the orange ALS and make it an “Other” type of ALS-wing. The NWr1 boxline does the damage.

The follow up brings the cluster to the state below.

where NW3 erased 2r1c2, turning 2r9c2 green and removing 1, and turning 2r9c7 blue, springing the “home with blue and seeing green” trap on 7r9c8.

From there, 6r9c4, 6r7c9 and 6r3c7 are trapped, 7r9c3 turns blue, removing 1r9c4, for a blxline 1r7c9, establishing SE7 and wrapping green in r7c4.

The fully expanded cluster places everything. This time the human oriented solvers of the review were stymied, until a simply conceived but arduous trial enabled one solver to reach a coloring solution to ultrahardcore 223.

Next week it’s Stefan Heine’s UHC 267. It may have only one post on the trial schedule, because a trial solution comes next week. You can win it more space by sending in your description of a trial free solution. But in case you don’t have one, here is the grid for the following review puzzle, Stefan Heine’s ultrahardcore 267.

Beeby notes the ANL isas a simple chain, but it’s AIC building class, with two cell wink nodes.

Then a relatively simple ALS_63,

is followed by the ALS_42, whose construction challenges drawing, and much moreso, spotting.

Now imagine spotting this double ALS_68, so called because there’s two restricted commons. Each ALS gives up a value, one 6, the other, 8. Other value sets in each ALS contain a true candidate.

The naked pair C49 implies W4, and a “discontinuous loop” takes another nibble. As in ANL, we can omit the weak link closure that breaks the AIC sequence,. The wink from 1r2c6 is direct, “seeing” in c6 when it is true.. The wink from 9r6c7 comes from the 1-way AIC when 1r2c6 is false. This chain is used again later, when intermediate removals make it more damaging.

Meanwhile Beeby’s “complex” chains get a workout. These have to be 1-way’s because the branches that make way for the chain exist in one direction. It is Phil Beeby’s baby. Humanly possible, under sufficient pressure.

Leaving from 1r4c6 one branch removes 1r9c4, and the other, 2r9c2, allowing the chain to confirm 1r9c2, if 1r4c6 is false, that is. Either 1r4c6 or 1r9c2 is true.

A second discontinuous 1-way nibble,

and another discontinuous 1-way,

and Beeby throws another complex curve ball. No need to trace this one out for anyone.

The complex removal enables the 2 chain ANL,

and the discontinuous 1-way from 5r8c5, whose 5r8c2 removal leaves a NW c1 boxline removing three 5s in NW, which picks up a 5-wing removing 5r9c4.

Now the hognose chain returns as an ANL, but with a second cluster. The clash of red and green in r4c2 means that orange or blue or both are true. The clusters share value 3. Any 3 seeing orange and blue is false.

Let’s save the finish of this overlong path for next week. If you go back to the last clue, you can run Beeby(philsfolly.com) and see how the action looks in text form.

]]>At the halfway point of the review UHC 223 starts tough, conceding four clues in basic.

Those 6f: to 8f: lines and their correspondingly long fill strings map onto a severely crowded grid.

Sudokuwiki leads off with two of its digit forcing chains, first a group mittened AIC ANL,

then a digit forcing chain boomer from 4r4c1. Then, being restricted from using cell or unit forcing chains, Sudokuwiki runs out of options.

Similarly, Beeby dries Similarly, Beeby dries up after this unlikely ALS_68.

There is a ray of hope in a nearby Single Allternate Sue de Coq, namely

NEc2 = 4(2+3)(6+9) +649.

The bv23 match in both remainders brings a clue and 8 candidate remainders. Having slogged through the review this far, you know what comes next.

It’s a trial of the second term 649, which adopts blue and gathers enough clues to get the solvers working.

The trial continues with this grouped ANL, and a band of helpers.

First an XY ANL (black) removes 1r9c2, creating the naked pair SE25.

Then the ANL is extended to make the terminals 8 (red), removing 8r7c3.

Then the ANL is extended to make the terminals 8 (red), removing 8r7c3.

And of course, a BARN.

The beat goes on with another XY ANL,

and after a brief follow up,

yet another one.

Trials don’t normally require a series of AIC eliminations, but the ultrahardcore are not normal.

It takes a bit more follow up to reach a contradiction, a second 2 in the NW box.

Since NEr2 = 649 fails,

NEr2 = 4(2+3)(6+9) . The (6+9) alternate comes from r2c2; the (2+3)alternate, from r2c4 and r1c8.

Next week, we go on without these 7 candidates to a solution to 223. If you already have your trial free solution, then after sending in your guidepost description to have it published here, you may want to get started with ultrahardcore 267.

tef

]]>Going back two posts, here is the full grid before the trial. So happens, there’s an empty BARN that Sudokuwiki did not report, because there are no victims.

The follow up to the new clues 8r2c9 and 8r5c8 includes a boxline elimination.

In the unique rectangle type 1, Beeby removes 7r1c6, but we can also remove 5r1c6. Either 5 or 7 being true generates a deadly rectangle.

Beeby continues with the ALS_43 below

and the overlapped ALS-wing to the right.

The complex 1-way takes less time than a regular AIC on Beeby. I think it’s because the full use of one way AIC makes much more of each opportunity, and more are available. When you ask for one, you get the first one Beeby comes to. This one starts on 5r1c9, suppresses two 5’s to make 5-slink in the SW box. It looks like an ANL, but is not. It depends on the 1-way AIC. If the starting value is true, both victims are false, and if it is false the AIC turns on 5r3c4, and both are false.

There’s much the same structure in the second complex 1-way. There’s some shifting because we’ve added a cluster and added slinks.

Coloring brings a trap in r1c3.

Next is the ALS_36, with an aligned 6 group,

and an overlapped ALS_57.

One more complex 1-way, requiring only one stiff arm block to get the 1-way AIC to 9r5c4, which if false regardless of 1r6c6 being true or false. The Cr6 boxline removes another 1.

Finally a chain ANL prompts us to look for a second cluster, since the blue/green one isn’t growing.

Interestingly:

- it has blue/green and red/orange 5’s and
- there’s a bridge. Blue and red together in r1c7 mean that blue and red are not both true. So either green or orange or both is true.

All of the candidates of the true pattern are the same color. For any combination of two colors to be true the two colors must eventually merge.

In the 5-panel, the combination of green and orange is possible for only one pattern. That’s a very specific case to try.

Removing in order, blue. red, seeing 5, seeing green, seeing orange, the normal breadth first trace leaves 6,9 in 3 cells of c6.

The combination of green and red leaves four patterns available. The trial is less specific.

The contradiction, traced below, is unusual.

9r5c5 forces a hidden pair in the West box, to join with NW79 for a deadly 79 rectangle.

The trials leave blue and orange freeforms to define possible 5 patterns.

On the grid, the 5 pattern restrictions produce a naked quad and naked quint placing N3.

That’s followed by

and a hidden unique rectangle as 9r4c6 forces 6 in two corners, and the slink forces 9 in the opposite corner.

The contest is ends with a Beeby ALS_46, and the SWr9 boxline triggering the collapse with S9.

Next week, its on to ultrahardcore 223.

]]>

My 5-panel patterns are much less promising than the 8-panel ones.

Too many patterns in both sets.

On the 8-panel there were two ways to start in the first 2 columns, while the 5-panel has three. Or to put it a better way, on the 8-panel it was 2 clues vs 2 other clues, giving us 2 clues either way. Remember how two clues were enough to get us past solver stalls on UHC 135?

Well we could have decided that with a lot less work. One thing you get with a full analysis is the orphans, the candidates not on any pattern. Do we have any on the 5-panels? I think I have one. All 5 pattern freeforms start in r1c9 or r9c9, and none include 5r9c2. Not much help.

Back to the plan, let’s see if we can get to coloring with two clues from the 8-panel value pattern analysis. It’s one trial to eliminate three or four more. We start by restarting at the stall point with the givens, plus one clue from the bypass, and now with two clues from one of the 8-pattern sets, 8r2c9 and 8r5c8.

Following up (NE8, E8) with NW5, Beeby pulls in an ALS_75, and we’re off again, but in trial mode.

Next is an ANL with an ALS value set on one terminal. The victim must see the whole value set.

Beeby follows with another hat, even more extreme. The 2 value set is crosswise, but there’s still a victim, and it comes with a clue.

We get NE1, and NW1m and W14, which show up here with a hidden UR, type 2.

Next, a couple of Beeby specials, down the middle.

First, a discontinuous 1-way: 5r5c4 is false if 5r3c4 is true, or if it is false, and the slink chain forces it out with a 4 in r5c4. Beeby notes call it a discontinuous loop.Put in the winks, and there is a loop. It’s the alternation of inferences that is discontinuous.

The other special is a complex 1-way. *Complex* 1-ways allow slink partners, true when 3r1c6 is false, erase candidates interfere with the AIC. You don’t search for these. You just build them with the hope that something will happen. And often enough, it does.

Now a very similar pair of 1-ways, another force out on the left, and a complex ANL on the right. It’s not a regular ANL, because the chain is 1-way. Two bonus removals are a NWc4 boxline, for a naked pair r2s79, which removes 9r2c4.

Remember, all of this in a trial to see if it’s this or another pair of clues, fits the solution.

Anyway, spot two long ALS in parallel lines with two matching singles and you might have this ALS_64.

In the same neighborhood, a hidden unique rectangle. 3r8c5 would force 2 in two corners, and the slink would force a 3 in the opposite corner.

The examples keep teaching. Next is an ALS 43 in which both Z sets are not singles, and only one is aligned with the victim. The other one is in the box with it.

Next is a simple AIC ANL whose removal triggers a productive naked pair.

The hidden UR is a direct result, but the removal from a crowded cell is slow progress.

Finally, it happens. An ANL brings a naked pair in c9, in turn bringing E2. Coloring is on, and traps leave a single 9 in c4, wrapping green.

The blue army then reaches a contradiction, when it removes the last 9 from W and from c3.

In the breadth first trace below, the blue army then reaches a contradiction,

when it removes the last 9 from W and from c3.

The big picture summary: We found that the UHC 179 8-patterns allowed only two placements of two 8’s in columns c89. We now know that one of those placements, and all the patterns including it, are false. We can put in the two clues of the other placement, go for a solution next week. And you’re invited.

Could it be time for your next attempt to get there first with the second documented trial free solution of the Stefan Heine ultrahardcore right page Sysudoku review? If you get one, I’ll draw the grids and interpret your moves and guess at how you spotted them for Sysudoku readers, and you can order a box of copies to hand out to your friends, relatives and other admirers. Here’s the next one, ultrahardcore 223.

]]>